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Let us Study

• Bernoulli Trial

• Binomial distribution

• Mean and variance of Binomial Distribution.

Let us Recall

• Many experiments are dichotomous in nature. For example, a tossed coin shows a ‘head’ or ‘tail’,

A result of student ‘pass’ or ‘fail’, a manufactured item can be ‘defective’ or ‘non-defective’, the

response to a question might be ‘yes’ or ‘no’, an egg has ‘hatched’ or ‘not hatched’, the decision is

‘yes’ or ‘no’ etc. In such cases, it is customary to call one of the outcomes a ‘success’ and the other

‘not success’ or ‘failure’. For example, in tossing a coin, if the occurrence of the head is considered

a success, then occurrence of tail is a failure.

Let us Learn

8.1.1 Bernoulli Trial :

Each time we toss a coin or roll a die or perform any other experiment, we call it a trial. If a coin

is tossed, say, 4 times, the number of trials is 4, each having exactly two outcomes, namely, success or

failure. The outcome of any trial is independent of the outcome of any other trial. In each of such trials,

the probability of success or failure remains constant. Such independent trials which have only two

outcomes usually referred to as ‘success’ or ‘failure’ are called Bernoulli trials.

Definition:

Trials of a random experiment are called Bernoulli trials, if they satisfy the following conditions :

(i) Each trial has exactly two outcomes : success or failure.

(ii) The probability of success remains the same in each trial.

Throwing a die 50 times is a case of 50 Bernoulli trials, in which each trial results in success (say

an even number) or failure (an odd number) and the probability of success ( p) is same for all 50

throws. Obviously, the successive throws of the die are independent trials. If the die is fair and has

six numbers 1 to 6 written on six faces, then

p =

1

2

and q = 1 − p ∴ q =

1

2

8. BINOMIAL DISTRIBUTION

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For example :

Consider a die to be thrown 20 times. if the result is an even number, consider it a success, else it is

a failure. Then p =

1

2 as there are 3 even numbers in the possible outcomes.

If in the same experiment, we consider the result a success if it is a multiple of 3, then p =

1

3

as there

are 2 multiples of 3 among the six possible outcomes. Both above trials are Bernoulli trials.

SOLVED EXAMPLE

Ex. 1 : Six balls are drawn successively from an urn containing 7 red and 9 black balls. Tell whether

or not the trials of drawing balls are Bernoulli trials when after each draw the ball drawn is

(i) replaced (ii) not replaced in the urn.

Solution :

(i) The number of trials is finite. When the drawing is done with replacement, the probability of

success (say, red ball) is p =

7

16 which is same for all six trials (draws). Hence, the drawing of

balls with replacements are Bernoulli trials.

(ii) When the drawing is done without replacement, the probability of success (i.e. red ball) in first

trial is 7

16

in second trial is 6

15

if first ball drawn is red and is 7

15

if first ball drawn is black

and so on. Clearly probability of success is not same for all trials, hence the trials are not

Bernoulli trials.

8.2 Binomial distribution:

Consider the experiment of tossing a coin in which each trial results in success (say, heads) or

failure (tails). Let S and F denote respectively success and failure in each trial. Suppose we are interested

in finding the ways in which we have one success in six trials. Clearly, six different cases are there as

listed below:

SFFFFF, FSFFFF, FFSFFF, FFFSFF, FFFFSF, FFFFFS.

Similarly, two successes and four failures can have 6!

4 ! × 2 !

= 15 combinations.

But as n grows large, the calculation can be lengthy. To avoid this the number for certain probabilities

can be obtained with Bernoullis formula.For this purpose, let us take the experiment made up of three

Bernoulli trials with probabilities p and q = 1 – p for success and failure respectively in each trial. The

sample space of the experiment is the set

S = �SSS, SSF, SFS, FSS, SFF, FSF, FFS, FFF�

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The number of successes is a random variable X and can take values 0, 1, 2, or 3.The probability

distribution of the number of successes is as below :

P (X = 0) = P (no success)

= P (�FFF�) = P(F )·P(F )·P(F ) , since trials are independent.

= q · q · q = q3

P (X = 1) = P (one success)

= P (�SFF, FSF, FFS�)

= P (�SFF�) + P (�FSF�) + P (�FFS�)

= P (S)·P(F )·P(F) + P(F)·P(S)·P(F) + P(F)·P(F)·P(S)

= p·q·q + q·p·q + q·q·p = 3pq2

P (X = 2) = P (two success)

= P (�SSF, SFS, FSS�)

= P (�SSF�) + P (�SFS�) + P (�FSS�)

= P(S)·P(S)·P(F) + P(S)·P(F)·P(S) + P(F)·P(S)·P(S)

= p·p·q + p·q·p + q·p·p = 3p2

q

and P (X = 3) = P (three successes)

= P (�SSS�)

= P(S)·P(S)·P(S)

= p3

Thus, the probability distribution of X is

X 0 1 2 3

P (X ) q3 3q2

p 3qp2 p3

Also, the binominal expansion of

(q + p)3

is q3

+ 3q2 p + 3 qp2

+ p3

Note that the probabilities of 0, 1, 2 or 3 successes are respectively the 1st, 2nd, 3rd and 4th term in the

expansion of (q + p)

3

.

Also, since q + p = 1, it follows that the sum of these probabilities, as expected, is 1.Thus, we may

conclude that in an experiment of n-Bernoulli trials, the probabilities of 0, 1, 2,..., n successes can be